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02 May 2021, 22:08
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Sum will be number of terms times average of all values
number of terms = 256
each digit can take one of 4 integers, hence 4*4*4*4
the average number will be (min+max)/2 = (1111+4444)/2 = 2,777.5
2,777.5 * 256 = 711,040
number of terms = 256
each digit can take one of 4 integers, hence 4*4*4*4
the average number will be (min+max)/2 = (1111+4444)/2 = 2,777.5
2,777.5 * 256 = 711,040
01 Oct 2021, 05:00
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Bunuel
Hi Bunuel, can you explain the logical way like you did for when the repetition is allowed for when the repetition is not allowed?
I am unable to figure it out. Just curious
01 Oct 2021, 14:34
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Hi ak2121,
You can calculate that total in the same general way - it's just that there wouldn't be as many overall numbers to consider. Here's how:
When forming 4-digit numbers with the digits 1, 2, 3 and 4 - but with NO repetitions - we have (4)(3)(2)(1) = 24 possible numbers. Since each digit can be in any of the 4 'spots', this means that we'll have...
6 numbers that begin with a 4
6 numbers that begin with a 3
6 numbers that begin with a 2
6 numbers that begin with a 1
That will hold true for each of the digits in the 4-digit number, so the overall sum of all those thousands, hundreds, tens and units digits would be...
6(4+3+2+1)(1000) + 6(4+3+2+1)(100) + 6(4+3+2+1)(10) + 6(4+3+2+1)(1)
You can actually simplify this in advance if you realize that 4+3+2+1 = 10...
6(10)(1000) + 6(10)(100) + 6(10)(10) + 6(10)(1) = 66,660
GMAT assassins aren't born, they're made,
Rich
You can calculate that total in the same general way - it's just that there wouldn't be as many overall numbers to consider. Here's how:
When forming 4-digit numbers with the digits 1, 2, 3 and 4 - but with NO repetitions - we have (4)(3)(2)(1) = 24 possible numbers. Since each digit can be in any of the 4 'spots', this means that we'll have...
6 numbers that begin with a 4
6 numbers that begin with a 3
6 numbers that begin with a 2
6 numbers that begin with a 1
That will hold true for each of the digits in the 4-digit number, so the overall sum of all those thousands, hundreds, tens and units digits would be...
6(4+3+2+1)(1000) + 6(4+3+2+1)(100) + 6(4+3+2+1)(10) + 6(4+3+2+1)(1)
You can actually simplify this in advance if you realize that 4+3+2+1 = 10...
6(10)(1000) + 6(10)(100) + 6(10)(10) + 6(10)(1) = 66,660
GMAT assassins aren't born, they're made,
Rich
